chapter 1

Chapter 14: The Analysis of Categorical Data11.a. The six intervals must be symmetric about 0, so denote the 4 th , 5 th and 6 th intervals by [0,1 1a0, [a, b), [b, ∞ ). a must be such that ( ) = . 6667( + )2 6gives a ≈ . 43 . Similarly Φ( b ) = . 8333 implies b ≈ . 97Φ a , which from Table A.3, so the six intervals are( − ∞ , -.97), [-.97, -.43), [-.43, 0), [0, .43), [.43, .97), and [.97, ∞ ).b. The six intervals are symmetric about the mean of .5. From a, the fourth interval shouldextend from the mean to .43 standard deviations above the mean, i.e., from .5 to .5 +.43(.002), which gives [.5, .50086). Thus the third interval is [.5 - .00086, .5) = [.49914,.5). Similarly, the upper endpoint of the fifth interval is .5 + .97(.002) = .50194, and thelower endpoint of the second interval is .5 - .00194 = .49806. The resulting intervals are( − ∞ , .49806), [.49806, .49914), [.49914, .5), [.5, .50086), [.50086, .50194), and[.50194, ∞ ).1c. Each expected count is 45( 6) = 7. 5 , and the observed counts are 13, 6, 6, 8, 7, and 5, soχ2 = 5.53. With 5 d.f., the p-value > .10, so we would fail to reject H o at any of theusual levels of significance. There is no evidence to suggest that the bolt diameters arenot normally distributed.Section 14.212.a. Let θ denote the probability of a male (as opposed to female) birth under the binomialmodel. The four cell probabilities (corresponding to x = 0, 1, 2, 3) are( θ ) = ( − ) 3, π ( ) ( ) 22θ = 3θ1 −θ, π ( θ ) = 3θ233( 1 −θ) , and ( θ )π11 θ3n 2 + n3n+ 2n+ n n + 2n+ 3n⋅ 1 −θ⋅θ31 2 3 2 3 4The likelihood is ( )π = .4θ. Forming the log likelihood,taking the derivative with respect to θ , equating to 0, and solving yieldsnθ =+ 2n3+ 3n66 + 128 + 48== .5043n480.5043 = 19. 480 .504 .496 59.ˆ 24160 12( − ) 52 , ( )( ) 5222⎡ ( 14 −19.52) ( 16 − 20.48)⎤. The estimated expected counts are2= , 60.48, and 20.48, soχ = ⎢+ ... +⎥ = 1.56 + .71 + .20 + .98 = 3.45 .⎣ 19.5220.48 ⎦The number of degrees of freedom for the test is 4 – 1 – 1 = 2. H o of a binomial2 2distribution will be rejected using significance level .05 if χ ≥ χ = 5. 992 .Because 3.45 < 5.992, H o is not rejected, and the binomial model is judged to be quiteplausible..05,2ˆ 53=150b. Now θ = . 353 and the estimated expected counts are 13.54, 22.17, 12.09, and2.20. The last estimated expected count is much less than 5, so the chi-squared test basedon 2 d.f. should not be used.443